Daily life examples of force

1. Daily life examples of force

2. 100 N 100 N 1 2 students are playing tug-of-war (拔河). 100 N 100 N If the rope is cut & reconnected by a newton balance, what is the reading? A 0 N C 200 N B 100 N

3. 2 An MTR train is leaving a station. Which compartment has the largest acceleration? A The one with fewest passengers. B The accelerations are different for different compartments. C All have the same acceleration.

4. 3 You step on a weighing scale and it reads 60 kg. What does it mean? A Your weight is 60 kg. B Your mass is 60 kg. C Your weight is 60 N.

5. 1 Tension in a string Tension – force in a string (depending on what it acts on) – either up/downwards consider forces acting on balance upward force by spring 10 N downward tension 10 N consider forces acting on mass upward tension10 N 1 kg weight 10 N

6. 1 Tension in a string Free-body diagram many forces to consider! To avoid confusion  we draw separate diagram  show forces acting on an object 1 kg

7. 1 Tension in a string Free-body diagram 1 Draw the body under consideration only. 1 kg 1 kg

8. 1 Tension in a string Free-body diagram 2 Identify all the forces acting on the body & use arrows to represent them. 1 kg 1 kg

9. 1 Tension in a string Free-body diagram 3 Label these forces on the body. upward tension10 N upward force by spring 10 N 1 kg 1 kg weight 10 N downward tension 10 N

10. 2 Normal force A book put on a table doesn't fall to the ground even it is acted by its weight. any force to balance the weight? normal force weight an upward force provided by the table... …acting perpendicular (normal) to the table Net force acting on book = 0

11. 2 Normal force Similarly, if you press the book against a wall, force by the hand normal force the wall provides a normalforce on the book to balance the applied force.

12. Q1 Two 4-kg masses… Two 4-kg masses connected by a string are hanging over a smooth pulley. They are stationary. What is the tension (T ) in the string? A 4 N B 8 N C 40 N D 80 N T 4 kg T 4 kg 40 N 40 N

13. Q2 3 dogs A, B & C are… T1 T2 N N f2 f1 W 3 dogs A, B and C are pulling a sledge at a uniform speed. A B C Draw the free-body diagram for dog B.

14. Q3 The force in a string is … tension The force in a string is called ________ . For a uniform and light string, this force at different points ______________________ (must be the same / can be different). must be the same

15. 3 Weight  the pull of gravity Weight is the gravitational force acting on an object by the Earth. Measured in newton (N) if an object falls freely, its weight Wgives it an acceleration of g. ApplyingF = ma Simulation W = mg

16. 3 Weight  the pull of gravity W = mg Taking g to be 10 m s2 the weight of a 1-kg mass is W = 1 kg  10 m s2 = 10 N

17. 3 Weight  the pull of gravity a Mass and weight Mass Weight gravitational force a measure of an object’s inertia change from place to place, depends on g same in everywhere SI unit: kg SI unit: N symbol: m symbol: W

18. 3 Weight  the pull of gravity g(m s2) g(m s2) Planet Planet 3.8 9.1 Mercury Saturn Venus 8.9 Uranus 8.8 9.8 (~10) 11.1 Earth Neptune 3.7 0.7 Mars Pluto Jupiter 24.3 a Mass and weight The acceleration due to gravity on the sufaces of planets:

19. 3 Weight  the pull of gravity a Mass and weight Video

20. 3 Weight  the pull of gravity m1 W1 = m2 W2 b Beam balance & spring balance The weights of 2 different masses, m1 & m2, in the same place: W1 = m1g & W2 = m2g Weights of different objects are proportional to their masses in a particular place.

21. 3 Weight  the pull of gravity b Beam balance & spring balance  beam balance can measure mass by comparing weight of an unknown mass with weight of known standard mass beam balance standard masses unknown mass

22. 3 Weight  the pull of gravity b Beam balance & spring balance A spring balance measures weight but NOT mass. It gives different values of weight of the same object at different places. spring balance

23. 3 Weight  the pull of gravity c Apparent change in weight in a lift When you travel in a fast lift, there are times when you can feel a loss or gain of weight. How does this happen?

24. 3 Weight  the pull of gravity c Apparent change in weight in a lift Video Simulation

25. W R c Apparent change in weight in a lift There are 2 forces acting on the girl: unchange 1 weight W& 2 normal force R by the lift floor – can be measured by a weighing scale – scale reading shows the weight felt by the girl, i.e. the girl's apparent weight may change

26. 10/F 10/F 10/F 10/F 10/F G/F G/F G/F G/F G/F c Apparent change in weight in a lift R changes when travelling in the lift. E.g. 10/F  G/F decelerates downwards R > W rest at G/F rest at 10/F R = W accelerates downwards R = W uniform velocity R < W R = W

27. W R c Apparent change in weight in a lift By Newton's 2nd law, the acceleration of the lift & the girl can be found if R is known. Video

28. Example 5 Mathew & his children are playing in a garden. Tony Joan Mathew Assume friction is negligible. Tony + his car = 17 kg Joan + her car =15 kg

29. Example 5 (a) Draw the free-body diagram for Tony & his car. Tony + his car = 17 kg Joan + her car =15 kg Tony Joan Mathew Assume friction is negligible. normal force pulling force by Mathew (P ) tension (T ) weight

30. Example 5 (b) Draw the free-body diagram for Joan & her car. Tony + his car = 17 kg Joan + her car =15 kg Tony Joan Mathew Assume friction is negligible. normal force tension weight

31. Example 5 (c) If the tension in the string connecting the 2 toy cars is 7.5 N, find Joan's acceleration. +ve Tony + his car = 17 kg Joan + her car =15 kg net force on Joan & car is the tension in string By F= ma, normal force 7.5 = 15×a tension a= 0.5 m s–2 to the right weight

32. Example 5 (d)Find Mathew's pulling force. Tony + his car = 17 kg Joan + her car =15 kg Tension (T ) = 7.5 N Joan's acceleration = 0.5 m s1 P T For Tony & his car, the net force is PT By F= ma P 7.5 = 17×0.5 P= 16 N

33. Example 5 (d)Find Mathew's pulling force. Tony + his car = 17 kg Joan + her car =15 kg Tension (T ) = 7.5 N Joan's acceleration = 0.5 m s1 P Alternatively, treat Tony, Joan & their cars as a single body. By F= ma P= (17 + 15) ×0.5 P= 16 N

34. Example 6 At lift-off, Saturn V rocket has an initialacceleration of 12 m s–2. Its mass is 3×105kg. What is the thrust (推力) of therocket?

35. Example 6 Thrust U Rocket's mass = 3  105 kg initial acceleration = 12 m s2 Applying the Newton’s 2nd law of motion, F= ma U– mg = ma U= ma+ mg = 3105 (12 + 10) = 6.6105N mg

36. 550 W 500 R 450 0 5 20 25 28 Example 7 mass of girl = 50 kg when lift is stationary on 10/F, reading of weighing scale = 500 N lift takes 25 s to travel to G/F scale reading/ N Find out the acceleration of the lift in a) 0-5 s, b) 20-25 s. time / s

37. 550 W 500 R 450 0 5 20 25 28 Example 7 mass of girl = 50 kg when lift is stationary on 10/F, reading of weighing scale = 500 N lift takes 25 s to travel to G/F +ve scale reading/ N (a) Take downwards as +ve During 0-5 s, R < W & W  R = ma  500  450 = 50a  a = 1 m s2 time / s lift accelerates downwards at 1 m s2

38. 550 W 500 R 450 0 5 20 25 28 Example 7 mass of girl = 50 kg when lift is stationary on 10/F, reading of weighing scale = 500 N lift takes 25 s to travel to G/F +ve scale reading/ N (b) During 20-25 s, W < R & W  R = ma  500  550 = 50a  a = 1 m s2 lift decelerates downwards at 1 m s2 time / s

39. Q1 The unit of weight is… The unit of weight is A kg. B N. C kg or N depending on the weighing instrument.

40. Q2 What happens to the weight… What happens to the weight of an apple when it is falling towards the ground from a tree? A It decreases. B It keeps constant. C It increases.

41. Q3 A dog of mass 5 kg stands... A dog of mass5 kg stands inside a lift. When the lift moves downwards from rest, the force U acting on it by the lift floor is 45 N. Find the acceleration of the lift.

42. Q3 A dog of mass 5 kg stands... Weight of the dog = mg = ____ 50 N Take downwards as +ve When it accelerates downwards, (U/W ) < (U/W ) The lift accelerates __________________ (upwards/downwards) at _____ m s-2. (U/W )  (U/W ) =ma downwards _____  _____ = 5a 50 45 1  a= _____ m s-2 1